Problem Description
You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time). Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 10^9 + 7.
Key Insights
- The problem can be solved using Dynamic Programming (DP) to keep track of the number of ways to reach each index after a certain number of steps.
- Each position can be reached from the left, right, or by staying in the same position.
- The maximum index we need to consider is limited by the minimum of steps and arrLen to avoid unnecessary computations.
Space and Time Complexity
Time Complexity: O(steps * min(steps, arrLen))
Space Complexity: O(min(steps, arrLen))
Solution
To solve this problem, we use a dynamic programming approach where we maintain a 1D array dp that keeps track of the number of ways to reach each index after a certain number of steps. The state transition relies on the fact that to get to index i after s steps, we can come from:
i-1(moving right from the left)i+1(moving left from the right)i(staying in the same position)
We'll iterate through each step, updating the number of ways to reach each index by summing the possible ways from the neighboring indices, while ensuring we don't go out of bounds. Finally, the answer will be found in dp[0] after processing all steps.